Prove the following statement by mathematical induction. For every integer n ≥ 0, 7" - 1 is divisible by 6. 2 Proof (by mathematical induction): Let P(n) be the following sentence. 7" - 1 is divisible by 6. We will show that P(n) is true for every integer n ≥ 0. Show that P(0) is true: Select P(0) from the choices below. 061(7⁰-1) O 6 is a multiple of 7⁰ - 1 O 1 is a factor of 70 - 1 O (7⁰1) 16 The truth of the selected statement follows from the definition of divisibility and the fact that 70 - 1 = Show that for each integer k ≥ 0, if P(k) is true, then P(k+ 1) is true: Let k be any integer with k ≥ 0, and suppose that P(k) is true. Select P(K) from the choices below. O 6 is a multiple of 7k - 1 O 1 is a factor of 7k - 1 O (7k - 1) is divisible by 6 O 6 is divisible by (7k 1) [This is P(k), the inductive hypothesis.] We must show that P(k+ 1) is true. Select P(k+ 1) from the choices below. O 6 is a multiple of 7k + 1 - 1 O 1 is a factor of 7k +1 -1 O (7k+11) is divisible by 6 O 6 is divisible by (7k+1-1) By the inductive hypothesis and the definition of divisibility, there exists an integer r such that 7k- 1 = 6r, and so 7k = 6r + 1. Now 7k+117k. 7-1. When 6r + 1 is substituted for 7K in the above equation, and the right-hand side is simplified, the result can be expressed in terms of k and r as follows. 7k+1 16. - This quantity is an integer because k and r are integers. Select the final sentence from the choices below. O Hence, 1 is a factor of (7k+1-1), and so P(k+1) is true, which completes the inductive step. O Hence, 1 is a factor of (7k+11), and so P(+1) is false, which completes the inductive step. O Hence, (7k+11) is divisible by 6, and so P(k+1) is false, which completes the inductive step. O Hence, 6 is a multiple of (7k+1 - 1), and so P(k+1) is false, which completes the inductive step. O Hence, 6 is divisible by (7k+1-1), and so P(k+1) is true, which completes the inductive step. O Hence, (7k+1 - 1) is divisible by 6, and so P(k+1) is true, which completes the inductive step. [Thus both the basis and the inductive steps have been proven, and so the proof by mathematical induction is complete.]